Is x+y less than 1? (1) x is less than 8/9 (2) y is less

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Is x+y less than 1? (1) x is less than 8/9 (2) y is less [#permalink]

25 Jan 2008, 05:07

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Is x+y less than 1?

(1) x is less than 8/9

(2) y is less than 1/8

Please explain your answer

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Re: DS: Is x+y less than 1? [#permalink]

25 Jan 2008, 05:19

1 and 2 by themselves are insufficient since they do not tell us anything about y and x respectively.

1 and 2 together:

lets try extremes , 8/9 + 1/8 = 73/72 > 1

therefore insuff.

E

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Re: DS: Is x+y less than 1? [#permalink]

25 Jan 2008, 06:21

The OA is E. But I honestly didn't understand your extremes. Sure, I know what you did, but I can't see how those 2 can qualify as extremes. The answers could be negative too, but certainly not those 2 numbers since the statements have mentioned less than these 2 numbers. Just because those extreme numbers can equal to more than 1 doesn't mean that the smaller fractions would do the same. After all, those extreme numbers that you have chosen aren't suppose to be part of the calculations. maybe i've lost your track here, but can you please explain? thanks

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Re: DS: Is x+y less than 1? [#permalink]

25 Jan 2008, 07:04

tarek99 wrote:

Is x+y less than 1?

(1) x is less than 8/9

(2) y is less than 1/8

Please explain your answer

Give the fractions the same denominator

X < 8/9
$formdata=\frac{8}{9}+=+\frac{64}{72}$

Y < 1/8
$formdata=\frac{1}{8}+=+\frac{9}{72}$

Test it out using this same denominator. Remember that X must be LESS than 64/72 and Y must be LESS than 9/72. Using 72 as a denominator these are the largest fractions we can use unless we expand out from integers (and get into 63.99/72 sort of thing)
$formdata=\frac{63}{72}+\frac{8}{72}=\frac{71}{72}$

Now many people might stop here and say that C is sufficient. Let's use a larger denominator so we can do more precise tuning and see what happens.

X < $formdata=\frac{640}{720}$

and

Y < $formdata=\frac{90}{720}$

These fractions are the same as the one above, we just multiplied the top and bottom by 10. Now let's take the largest fractions we can get and add them together like we did above:

$formdata=\frac{639}{720}+\frac{89}{720}=\frac{728}{720}$

As you can see, this fraction is larger than 1. Making both statements together insufficient.

Answer E

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Re: DS: Is x+y less than 1? [#permalink]

25 Jan 2008, 07:33

tarek99 wrote:

ok since 1) says x<8/9 lets assume x=\frac{799}{900}(since its less than \frac{800}{900})

and lets assume that y=\frac{99}{800}

add these two together we get a number slightly more than 1. Try it

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Re: DS: Is x+y less than 1? [#permalink]

25 Jan 2008, 07:34

tarek99 wrote:

try x = 8/9 - 1/(72*3), y = 1/8 - 1/(72*3)

x+y = (64-1/3+9-1/3)/72 = (73-2/3)/72 = (72+1/3)/72 = 1 + 1/(3*72) > 1

then try x = 0, y = 0: x+y = 0 < 1

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Re: DS: Is x+y less than 1? [#permalink]

25 Jan 2008, 10:09

Using both statements, it is important to know the following:

If x+y=8/9 + 1/8 =73/ 72, it is more than 1. Then Suf

But you can think of x as a negative number and y as a negative number.
Then x+y will be negative, that is <1.

For this reason, the OA must be E.

Re: DS: Is x+y less than 1? [#permalink] 25 Jan 2008, 10:09

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Is x+y less than 1? (1) x is less than 8/9 (2) y is less

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Is x+y less than 1? (1) x is less than 8/9 (2) y is less

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